Gr. 12 Data Management

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In how many of the arrangements would the two L's be together, for the word BASKETBALL?

  • Gr. 12 Data Management -

    The number of arrangements for n distinct lettes is
    n!
    For example: 4!=24 distinct words can be made from the letters ABCD.

    The number of arrangements for n letters, of which p are identical is
    n!/p!
    For example: 4!/2!=12 words can be made from the letters AABC.

    The number of arrangements for n letters, of which p are identical and q are identical is
    n!/(p!q!)
    For example: 6!/(2!2!)=180 words can be made from the letters AABBCD.

    If two letters have to be together all the timee, treat them as a single letter.

    For the word BASKETBALL,
    first arrange them in alphabetical order:
    AABBEKLLST
    Out of the 10 letters, AA and BB are repetitions, LL can be treated as one single letter (to have a total of 11 letters).
    So the number of distinct words possible is
    11!/(...)

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